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261FE-F2011

# 261FE-F2011 - t 1 Let C be the curve given by ﬁt...

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Unformatted text preview: t 1. Let C be the curve given by ﬁt) = (Ah/13¢, 5 —— 232) for t > 0. At What point does the tangent line to C at (4, 1, 4) intersect the my plane? A. (0,1,0) B. (AREA/53,0) 0. (2,1,0) D. (8,3,0) E. (0,—1,0) ' 2. The arclength of the curve 770:) 2: 25+ tzi—i- (11173)]; for 2 g t g 4 is: (l 17 I . 4 + In 2 16 +1112 15 4 12+1n2 A. vibrato 3. A particle moves in space with acceleration EL’(t) m at}; and initial velocity and poSition given by 17(0) : 5, 77(0) :2 3+ Is. Where is the particle at time t = 2? A. (1,1,82) B. (0,1,e2) C. (0,1,3—1) D. (1,1,e2—2) E. (0,1,62—2) 4. Suppose that z is deﬁned implicitly as a function of a: and y by the equation What is the value of Q at 8m 6” + sin(7ryz) — myz = 0. (6,1,1)? A. -l e B. l e C, '“1 7r D, l 71' E. 1 eﬁ'rr E i i f i E E 5. The surface area of a rectangular box is given by the function 3(32, 3/, z) = 22:31 + Zyz + 2932 where x,y, z are its sides. These are measured as so : 10 cm, :1; = 20 cm, 2 a 30 cm with possible errors in measurements as much as 0.1 cm. Use differentials to estimate the maximum error in the calculated surface area. 12 cm2 24 0mg 36 cm2 48 cm2 60 01112 EbQW? 6. Given 55 == (1,-1,2> perpendicular to (i x 7. The intersection of the hyperbolic paraboloid m2 — y2 — z — 1 = 0 with the yz~p1ane consists of 0&9? A. nd 5: (2,1, 0), ﬁnd 1% such that the vector 5’: (5,1: — 1,2) is A. i=1 B. t:2 C. t=—1 D. t=——2 E. i=0 a hyperbola and a parabola . a hyperbola . an ellipse B C D. E two lines . a parabola 8. Let f(m, y) = x/mi’ + y. The equation for the tangent plane to z = f(a:,y) at (2,1) is 9, The critical points of f(a:, y) = 33:3 + 3y3 + m3y3 are: A. 2x/gz—4m—yzl B. 2x/Ez—4cc—y210 ‘ C. 2z—2m—y=1 D. 23—2m—y=10 E. 2x/gz—2w—y=9 A. (0,0),(1,—1) B. (0,0) C. (1,1) D. (0,0),(—31/3, —31/3) E. (—31/3,—31/3),(1,1) 10. The directional derivative of the function f (:0, y) = 43:31 + a” at the point (0,1) and in the direction of 17 = (3, —4) is: 11. If (92, then —— Is: at z: uz l +21 3 “(3)79 = t + 32: 11(3)” =1n(t) —(1 + %) (t+32)2+lnt ~(2<t+s2) + 9%) (t+s2)2 +lnt -(2(t+82) + %) W —(u+1) (UM-1))? ——(2u+ 1) 1&2 +1) ...
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